Problem: $x^2+xy+y^3=0$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{2x}{1+3y^2}$ (Choice B) B $-\dfrac{x+3y^2}{2x+y}$ (Choice C) C $-\dfrac{2x+y}{x+3y^2}$ (Choice D) D $-\dfrac{2x}{x+3y^2}$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $x^2+xy+y^3=0$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} x^2+xy+y^3&=0 \\\\ \dfrac{d}{dx}(x^2+xy+y^3)&=\dfrac{d}{dx}(0) \\\\ \dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(xy)+\dfrac{d}{dx}(y^3)&=0 \\\\ 2x+(1)(y)+(x)\cdot\dfrac{dy}{dx}+3y^2\cdot\dfrac{dy}{dx}&=0 \\\\ 2x+y+x\cdot\dfrac{dy}{dx}+3y^2\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 2x+y+x\cdot\dfrac{dy}{dx}+3y^2\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(x+3y^2)&=-(2x+y) \\\\ \dfrac{dy}{dx}&=-\dfrac{2x+y}{x+3y^2} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+3y^2}$.